Question: $\int \left( \dfrac{2 x^{3} +5 x^{2} -2 x}{x}\right)dx=$ $+C$
Answer: The integrand is the quotient of two functions: $2 x^{3} +5 x^{2} -2 x$ and $x$. Although it is tempting to take the quotient of their integrals, this would not work. $\int \dfrac{f(x)}{g(x)}\,dx\neq\dfrac{\int f(x)\,dx}{\int g(x)\,dx}$ Instead, what we should do is divide each term in the numerator by the denominator. $\begin{aligned} &\phantom{=}\int \left( \dfrac{2 x^{3} +5 x^{2} -2 x}{x}\right)dx \\\\ &=\int\left(\dfrac{2 x^{3}}{x}+\dfrac{5 x^{2}}{x}-\dfrac{2 x}{x}\right)dx \\\\ &=\int (2 x^{2} +5 x -2 )\,dx \end{aligned}$ Now we can integrate using the reverse power rule, the sum rule, and the constant multiple rule for indefinite integrals. $\begin{aligned} &\phantom{=}\int \left( \dfrac{2 x^{3} +5 x^{2} -2 x}{x}\right)dx \\\\ &=\int (2 x^{2} +5 x -2 )\,dx \\\\ &= 2\int x^{2}\,dx +5\int x\,dx -2\int1 \,dx \\\\ &=2\dfrac{x^3}{3} +5\dfrac{x^2}{2} -2\dfrac{x^1}{1}+C \\\\ &=\dfrac{2}{3} x^3 +\dfrac{5}{2} x^2 -2 x+C \end{aligned}$ In conclusion, $\int \left( \dfrac{2 x^{3} +5 x^{2} -2 x}{x}\right)dx=\dfrac{2}{3} x^3 +\dfrac{5}{2} x^2 -2 x+C$